RCBC Jnt SuperEle Model

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RCBC Jnt SuperEle Model

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  • Reinforced Concrete Beam-Column Joint: Macroscopic

    Super-element models

    -Nilanjan Mitra

    (work performed as a PhD student while at University of Washington between 2001-2006)

  • Need for the study

    Reinforced concrete beam column jointssubjected to earthquake loading

    Experimental Investigation@ UW

    I-280 Freeway, San Francisco, CAfollowing Loma Prieta Earthquake in 1989

    Courtesy: NISEE, Univ. of California, Berkeley.

  • Loading in a joint region

    Earthquake Loading of Beam-Column Joint

    compression resultant (concrete and steel)

    shear resultant (concrete)

    Earthquake Loading of Beam-Column Joint

    compression resultant (concrete and steel)

    compression resultant (concrete and steel)

    shear resultant (concrete)

    shear resultant (concrete) tension resultant (steel)

  • anchorage bond stress acting on joint core concrete

    compression force carried by joint core concrete

    Internal load distribution in a joint

  • Macroscopic beam-column joint element models

  • Macroscopic beam-column joint element models

  • Macroscopic beam-column joint element models

  • shear panel

    external node

    internal node

    rigid externalinterface plane

    shown with finite widthto facilitate discussion

    beam element

    zero-width region

    interface-shear spring

    bar-slip spring

    zero-length

    zero-length

    elem

    ent

    colu

    mn

    Proposed Beam-column super-element model 4-noded 12-dof element 8 bar-slip springs to simulate

    anchorage failure 4 interface-shear springs to simulate

    shear transfer failure at joint interface 1 shear-panel to simulate inelastic

    action of shear within joint core

    Note: The location of the bar-slipsprings is at the centroid of thetension-compression couple at nominalstrength of the beams.

    [Mitra & Lowes; J. Structural Eng. ASCE, 2007: 133 (1): 105-20 ]

  • Joint element formulation: Kinematics

    External, Internal and Component deformation

  • Joint element formulation: Equilibrium

    External, Internal and Component forces

    Solution of element state achieved by an iterative procedure and requires solving for zero reaction in the 4 internal degrees of freedom

  • Characterized by Response envelope Unload reload path Damage rules

    Hysteretic one dimensional material model

    deformation

    load

    (ePd1,ePf1)

    (ePd4,ePf4)

    (ePd3,ePf3)(ePd2,ePf2)

    (eNd3,eNf3)(eNd2,eNf2)

    (*,uForceP.ePf3)

    (dmin,f(dmin))

    (dmax,f(dmax))

    (rDispP.dmax,rForceP.f(dmax))

    (rDispN.dmin,rForceN.f(dmin))

    (*,uForceN.eNf3)

    (eNd1,eNf1)

    (eNd4,eNf4)

  • Damage simulation in material model

    -0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    deformation

    load

    without damagewith unloading stiffness damage

    -0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    deformationlo

    ad

    without damagewith reloading stiffness damage

    -0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    deformation

    load

    without damagewith strength damage

    ( ) ( )( )3 41 max 2i d = +

    max minmax

    max min

    max. ,i id d

    ddef def

    =

    ( ).f No of load cycles =

    ( )f Accumulated Energy =

    ( )0 1 ki ik k = ( ) ( ) ( )max max 0 1

    fii

    f f = ( ) ( ) ( )max max 0 1

    dii

    d d = +

  • Damage simulation in material model

    load historyi

    monotonic

    monotonic load history

    dEE

    EgE dE

    = =

    -0.015 -0.01 -0.005 0 0.005 0.01 0.015-8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    deformation

    load

    without damagewith all 3 damage rules

    -0.015 -0.01 -0.005 0 0.005 0.01 0.015-5

    -4

    -3

    -2

    -1

    0

    1

    2

    3

    4

    5

    deformation

    load

    with all 3 damages (Energy)with all 3 damages (Cyclic)

    max4duu

    =

    Energy criterion

    No. of load cycle criterion: rain-flow-counting algorithm

  • Shear-panel calibration

    column

    shear panel

    Shear panel envelope calibration MCFT Diagonal compression strut

    Compression envelope reduction

    Determination of hysteretic model parameters

    Typical response envelope

    Observed Simulated

    Spec

    imen

    SE8

    (S

    teve

    ns e

    t al.

    1987

    )

    -0.012 -0.008 -0.004 0 0.004 0.008 0.012-10

    -8

    -6

    -4

    -2

    0

    2

    4

    6

    8

    10

    Shear strain

    Shea

    r st

    ress

    (MPa

    )

  • Shear panel envelope calibration using proposedDiagonal compression strut mechanism

    _ cosc strut strut strutstrut

    jnt

    f ww

    =

    Mander et al. (1988) concrete

    Column longitudinal and joint hoop

    steel confine the strut.

    Reduction in concrete to account for

    perpendicular tensile stress to the strut

    cyclic loading.

    Strut force is converted to panel shear

    stress as

  • 2_

    _3.62 2.82 1 0.39

    0.45 0.39

    c strut t t t

    c Mander cc cc cc

    t

    cc

    ff

    = + 0

    Data with j = 0

    Vecchio 1986Stevens 1991Hsu 1995Noguchi 1992Proposed eq. for

    j > 0

    Proposed eq. for j = 0

    2_

    _0.36 0.60 1 0.83

    0.75 0.83

    c strut t t t

    c Mander cc cc cc

    t

    cc

    ff

    = + Eq. for Eq. for

  • Comparison of MCFT and Diagonal Compression Strut model in shear-panel envelope calibration

    0 0.5 1 1.5 2 2.50

    0.5

    1

    1.5

    2

    2.5

    mcf

    t_cy

    clic

    / m

    ax

    0.55JFBYJFBY

    0 0.5 1 1.5 2 2.50

    0.5

    1

    1.5

    2

    dia

    gona

    l_st

    rut /

    m

    ax

    JFBYJFBY

    [Lowes, Altoontash and Mitra, J. Structural Eng. ASCE, 2005: 131 (6) ]

    Transverse steel contribution to shear stress

  • Bar slip material model calibration

    column

    Bar-slip spring Mechanistic model :- envelope

    Hysteretic model calibration

    Strength deterioration model

    -2 0 2 4 6 8 10 12 14 16-1000

    -500

    0

    500

    1000

    slip (mm)

    bar-

    spri

    ng fo

    rce

    (kN

    )

    Typical response envelope

  • Bar slip mechanistic model

    Assumptions for anchorage response of bond within the joint region:

    Bond stress uniform for elastic reinforcement, piecewise uniform for reinforcement

    loaded beyond yield

    Slip is the relative movement of reinforcement bar with respect to the joint perimeter

    Slip is a function of strain distribution in the joint

    Bar exhibits zero slip at zero bar stress

    2

    0

    2fsl

    fsE b Eslip s y

    b b

    ldd x dx f fA E E d

    = =